All categories

3 years ago

What is 5/12 times 6 in multiply fractions with whole numbers

Mathematics

1 answer:

KiRa [710]3 years ago

3 0

5/12*6 =5/12*6/1 crosscancle you get 5/2 the you get 2 1/2

You might be interested in

3 56/34 - 5 1/2 simplify

MaRussiya [10]

Answer:

29/34

hope this helps

Step-by-step explanation:

4 0

3 years ago

okay, genuinely does anyone know how this work? if you do could you please explain how you get the answer-

fomenos

Answer:

The pdf is not showing anything

Step-by-step explanation:I would help you otherwise.I am sorry.Don't let this get you down.I hope you have a good day.

4 0

2 years ago

Can someone do this please?

amid [387]

Answer:

<1 =73

Step-by-step explanation:

The sum of the angles of a triangle add to 180 degrees

72+ 35 + <1 = 180

Add like terms

107 + <1 = 180

Subtract 107 from each side

<1 = 180-107

<1 =73

8 0

1 year ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection = </em>

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago

What is the solution to this system of linear equations?

svetoff [14.1K]

It’s B
If y-x=6
Y +X =_10
then y= 6 + x, instead of y insert this no
6+ x+ x =-10
6 + 2x =-10 then collect like terms
2x =-10-6
2x=-16 then multiple both side by 1/2
X=-8
Y=6+x instead of x insert -8
Y=6-8
=-2

7 0

3 years ago